A 5.0 mL sample of agua oxinada was diluted in 30 mL distilled water and 30 mL diluted sulfuric acid. The resulting solution was titrated wi

Question

A 5.0 mL sample of agua oxinada was diluted in 30 mL distilled water and 30 mL diluted sulfuric acid. The resulting solution was titrated with 1.4686N KMnO4 consuming 29.7mL to reach the endpoint. What is the % H2O2 content of the sample? Look for Hydrogen peroxide monograph and determine if the sample conforms to the specification.

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Iris 4 months 2022-01-03T04:44:18+00:00 1 Answer 0 views 0

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  1. Genesis
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    2022-01-03T04:45:38+00:00
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    Answer:

    14.8% of H2O2

    Explanation:

    Based on the redox reaction of H2O2/MnO4- is:

    2MnO4- + 5H2O2 + 6H+ ⇄ 2Mn2+ + 5O2 + 8H2O

    2 moles of KMnO4 react with 5 moles of H2O2

    To solve this question we need to find the moles of KMnO4 until reach the endpoint. Using the balanced reaction, we can find the moles of H2O2 that must be converted to grams using its molar mass -Molar mass H2O2: 34.015g/mol-:

    Moles KMnO4:

    0.0297L * (1.4686eq/L) = 0.04362eq MnO4- * (1mol / 5eq) = 0.008723 moles MnO4-

    Moles H2O2:

    0.008723 moles MnO4- * (5mol H2O2 / 2mol MnO4-) = 0.02181 moles H2O2

    Mass H2O2:

    0.02181 moles H2O2 * (34.015g/mol) = 0.7418g H2O2

    %:

    0.7418g H2O2 / 5.0mL * 100

    14.8% of H2O2

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45:5+15-5:5+20+17*12-15 = ? ( )